The two possible positions of an object kept in front of a lens of power +5.0 D, so that the image formed in both cases is four times magnified, are
(A) \( -30\,\mathrm{cm} \)
(B) \( -25\,\mathrm{cm} \)
(C) \( -15\,\mathrm{cm} \)
(D) \( -45\,\mathrm{cm} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (B) and (C) only
A convex lens (positive power) can form a magnified image in two ways: a real, inverted image with \( m = -4 \) and a virtual, erect image with \( m = +4 \). Use the lens formula and the magnification \( m = v/u \).
\( \displaystyle f = \frac{1}{P} = \frac{1}{5\,\mathrm{D}} = 0.2\,\mathrm{m} = 20\,\mathrm{cm} \)
\( \displaystyle \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)
Real image, \( m = -4 \), so \( v = -4u \):
\( \displaystyle -\frac{1}{4u} - \frac{1}{u} = \frac{1}{f} \;\Rightarrow\; -\frac{5}{4u} = \frac{1}{f} \;\Rightarrow\; u = -\frac{5f}{4} = -25\,\mathrm{cm} \)
Virtual image, \( m = +4 \), so \( v = 4u \):
\( \displaystyle \frac{1}{4u} - \frac{1}{u} = \frac{1}{f} \;\Rightarrow\; -\frac{3}{4u} = \frac{1}{f} \;\Rightarrow\; u = -\frac{3f}{4} = -15\,\mathrm{cm} \)
The two positions are \( -25\,\mathrm{cm} \) and \( -15\,\mathrm{cm} \), which are (B) and (C).